Let $\mathbf{X}$ be a stack over $Top$ (a lax sheaf of groupoids, or some such thing). If it admits a surjective representable map $F \to \mathbf{X}$ then one can form the iterated fibre product to get a simplicial space $F_\bullet$, and its realisation $X$ is the homotopy type of $\mathbf{X}$, which comes with a (homotopy class of?) map $X \to \mathbf{X}$.

It seems that the reasonable thing to call the singular cohomology of $\mathbf{X}$ is the singular cohomology of its homotopy type, because the homotopy type is sufficiently unique for this to be well-defined.

On the other hand, consider the composition of functors
$$\mathbf{h}^i : Top \overset{H^i(-;\mathbb{Z})}\to AbGp \to Set \overset{inc}\to Gpd$$
where the middle functor is forgetful, and the last one is the inclusion of sets as groupoids with no non-identity morphisms.

The stack $\mathbf{X}$ gives another functor $Top \to Gpd$, and one may consider the set $H^i$ of natural transformations $\eta : \mathbf{X} \to \mathbf{h}^i$. This set has the structure of an abelian group as one may add pointwise (as $\mathbf{h}^i$ factors through abelian groups).

Given any such $\eta$, we may apply it to the unique homotopy class of maps $X \to \mathbf{X}$ to obtain $\eta(X \to \mathbf{X}) \in H^i(X;\mathbf{Z})$ a cohomology class on the homotopy type. Any $Y \to \mathbf{X}$ factors up to homotopy through $X$, and so $\eta(Y \to \mathbf{X})$ is obtained by pullback from $\eta(X \to \mathbf{X})$. Thus it seems to me that the group $H^i$ is naturally isomorphic to $H^i(X;\mathbf{Z})$.

My first question, at last, is: is the group $H^i$ the correct notion of the cohomology of the stack $\mathbf{X}$, which happens to coincide with the cohomology of its homotopy type? Would it still be a reasonable definition on some terrible stack that does not admit an atlas?

Secondly: how does one do homology like this?

Thirdly: one can do the above over $Diff$ instead of $Top$, and use de Rham cohomology in the definition of $\mathbf{h}^i$. Then it seems the group one produces, call it $H^i_{dR}$ now, still makes sense, but the homotopy type of a stack over $Diff$ is not necessarily itself a manifold and does not necessarily have a de Rham theory. Is this a reasonable thing to do? Why is it not done this way (for example by Behrend)?

Just to make this clear to everyone, $F$ is assumed to be a topological space. Such a stack \mathbf{X} is equivalent to the stack of torsors of the topological groupoid object $X \times_{\mathbf{X}} X \rightrightarrows X$.
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David CarchediMay 12 '10 at 18:43

2 Answers
2

1.) is the group $\mathbf{H^{i}}$ the correct notion of the cohomology of the stack $\mathbf{X}$?

Yes, I believe so, at least when the stack is nice enough. However, you should be a little careful with your proof:

Behrang Noohi's develops a homotopy theory for topological stacks. In particular, every topological stack $\mathbf{X}$ admits an atlas $X \to \mathbf{X}$ which is a universal homotopy equivalence. (To be more precise, to any given any atlas $F \to \mathbf{X}$, there exists such a universal $X$- it is the classifying space of the topological groupoid associated to this atlas). Being a universal homotopy equivalence means that for any $Y \to \mathbf{X}$, the map $Y \times_{\mathbf{X}} X \to Y$ is a weak equivalence. This is the sense that any map $Y \to \mathbf{X}$ "factors through $X$ up to homotopy". Combining this observation with the naturality square for $\eta:\mathbf{X} \to h^{i}$ gives a precise proof of your statement. It should be noted that there is a wider class of stacks than just topological ones which admit such a universal weak equivalence from a space. For example, paratopological and pseudotopological stacks (see: Homotopy Theory of Topological stacks by Behrang Noohi).

Morevoer, in Gepner and Henrique's "Homotopy Theory of Orbispaces", they remark that any stack over $Top$ has a weak homotopy type, not just ones with atlases. So, we get a unique up to homotopy map from a CW-complex $X$ to $\mathbf{X}$ which is homotopy terminal amongst maps from a CW-complex. I do not know whether or not this is a universal weak equivalence in any way though. However, there might be a way of using this to extend the proof to show your definition agrees with the cohomology groups of this weak homotopy type.

Just for some extra knowledge, in the above paper, they also show that any map $Y \to \mathbf{X}$ factors DIRECTLY through $X \to \mathbf{X}$ (up to 2-iso) whenever $Y$ is paracompact Hausdorff. This is a really cool result.

2.) Well, you can always do homology by using this universal weak equivalence, but, this is probably not the kind of answer you are looking for.

3.) For differentiable stacks, if by differentiable stack you mean one coming from an atlas, you can always define De Rham cohomology via a double complex associated to the associated simplicial manifold arising as the nerve of any Lie groupoid representing you. See for instance "Differentiable Stacks and Gerbes" by Ping Xu. (There are a few other places I've seen this as well, I can search if you need me to).

4.) This is certainly the first time I myself have seen the cohomology defined this way, but, perhaps others have seen this. Anyhow, here is I would have a look at the three papers I've mentioned. All three are on the Arxiv.

Sure. While the homotopy type of a stack in $Diff$ doesn't exactly have a de Rham complex, it does have a couple of things that work perfectly as substitutes. Behrend uses the Cech-de Rham double complex from a particular choice of atlas. The homotopy type of the stack is the geometric realisation of a simplicial manifold, and the Cech-de Rham double complex is made from the simplicial object in cochain complexes by taking the de Rham complex levelwise, so it really is the de Rham model for the simplicial manifold.

Alternatively, the homotopy type of the stack has a real homotopy type (which can be defined, for example, by taking Sullivan's PL forms with real coefficients). On manifolds the cohomology of the real homotopy type is of course naturally isomorphic to the de Rham cohomology. Thus on stacks over $Diff$ the Cech-de Rham complex is a model for the real homotopy type.

What about K-theory?

This wasn't one of your questions, but as long as we are varying the cohomology theory, it's worth mentioning this. The K-theory of a stack really is different from the K-theory of its homotopy type. The simplest example is the stack $*/G$. The K-theory of the stack is the G-equivariant K-theory of a point, but K-theory of the homotopy type is K(BG), which is of course the completion at the augmentation ideal by the Atiyah-Segal theorem.

I think that defining K-theory in terms of natural transformations $hom(-,\textbf{X}) \to K(-)$ just gives you the K-theory of the homotopy type.

Hi Jeff. Does this double complex exist for a general stack on Diff? I'd suspect you'd need an atlas- I'm not sure how to get a simplicial manifold otherwise...
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David CarchediMay 13 '10 at 2:54

I think you are right - one needs an atlas to make a cech-de rham complex. If you don't have an atlas then you could try doing something along the lines of how you might construct a homotopy type for a stack without an atlas: consider the category of all spaces mapping to the stack. This gives a diagram of simplicial manifolds, and the homotopy type should be the hocolim I think. To get the analogue of the Cech-de Rham complex, you should probably take the (ho)lim of the corresponding diagram of complexes.
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Jeffrey GiansiracusaMay 13 '10 at 6:30

I'm curious, do you have a good reference showing how equivariant K-theory is the correct notion of K-theory on a stack? And does this extend to stacks coming from groupoids?
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David CarchediMay 14 '10 at 13:39